Final answer:
To find the initial height of a tire swing from which a tire is released and reaches a speed of 2.5 m/s at the bottom, we use the conservation of mechanical energy. The initial potential energy is equal to the kinetic energy at the bottom, which gives us an initial height of approximately 0.32 meters. Therefore, the correct answer is C.
Step-by-step explanation:
The problem described relates to the conservation of mechanical energy where the initial potential energy of the tire is converted into kinetic energy at the bottom of the swing. Since there is no mention of friction or any other forces doing work, we can assume that the mechanical energy is conserved. To find the initial height, we use the formula for gravitational potential energy PE = mgh and set it equal to the kinetic energy KE = 1/2mv^2 at the lowest point of the swing where m is the mass of the tire, g is the acceleration due to gravity, and v is the final velocity.
The mass of the tire cancels out in our calculations, so we can solve for the initial height h using:
mgh = 1/2mv^2
gh = 1/2v^2
h = (1/2v^2)/g
Substituting the given values:
h = (1/2×(2.5)^2) / 9.8
h ≈ 0.32m
Therefore, the initial height from which the tire was released is about 0.32 meters.